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Question 99 of 148

Q.The value of 'c' in Rolle's Theorem for the function f(x)=cos⁡x2f(x) = \cos\dfrac{x}{2} on [π,3π][\pi, 3\pi] is :

(a) 0
(b) 2π2\pi
(c) π2\dfrac{\pi}{2}
(d) 3π2\dfrac{3\pi}{2}
Puducherry TnboardTamil Nadu HSC (DGE) Board 2017MCQ· 1mImportance★★★★★
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Verify Rolle's hypotheses on [π,3π][\pi,3\pi] and solve f′(c)=0f'(c)=0; the only root of sin⁡(c/2)=0\sin(c/2)=0 lying strictly between π\pi and 3π3\pi is c=2πc=2\pi.

  1. f(x)=cos⁡x2f(x)=\cos\dfrac{x}{2} is continuous on [π,3π][\pi,3\pi] and differentiable on (π,3π)(\pi,3\pi) (cosine is differentiable everywhere), so Rolle's theorem applies provided the end values match.
  2. Check end values: f(π)=cos⁡π2=0f(\pi)=\cos\dfrac{\pi}{2}=0 and f(3π)=cos⁡3π2=0f(3\pi)=\cos\dfrac{3\pi}{2}=0. So f(π)=f(3π)f(\pi)=f(3\pi), and Rolle's theorem guarantees at least one c∈(π,3π)c\in(\pi,3\pi) with f′(c)=0f'(c)=0.
  3. Differentiate: f′(x)=−12sin⁡x2f'(x) = -\dfrac{1}{2}\sin\dfrac{x}{2}. …

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